The first equilibrium condition requires that the vector sum of all the forces acting on a body be zero, so that it is at rest (static equilibrium) or with uniform rectilinear motion (dynamic equilibrium).
This sum of forces is none other than the net force that acts on the body, expressed mathematically in this way:
Fnet = 0
∑ F = 0
In space, the first equilibrium condition gives rise to three equations, one for each dimension:
∑ Fx = 0; ∑ FY = 0 and ∑ Fz = 0
When these equations are satisfied, the object does not translate or if it does, it will be with constant velocity.
Looking around us we realize that we continually seek to satisfy the first condition of balance so that things do not fall apart..
For this reason, it seeks to compensate the gravitational attraction of the Earth by means of supports, ropes or supports of some, so that in this way things remain in their place and do not go to the ground..
Other times what is needed is to prevent external electromagnetic fields from interfering with the operation of electrical circuits and communication devices. In this case, it is the electric charges that must be in equilibrium..
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A large number of everyday objects satisfy the first condition of equilibrium, it is a matter of carefully observing:
Builders seek stability in constructions so that users remain safe. The objective of statics is to study the conditions for static equilibrium to occur in buildings, bridges, roads and all kinds of structures..
These signaling devices must remain fixed to fulfill their functions, therefore they are held by cables, poles and rods in such a way that the first equilibrium condition is met..
When conductive materials such as copper and other metals acquire an electrical charge, electrostatic equilibrium is soon established, leaving the excess charge on the conductive surface. Inside the electric field is zero.
This effect is often used to isolate electrical and electronic equipment from external fields, using the so-called Faraday cage. The cage is made of conductive material and surrounds the equipment to be protected.
During storms, automobiles serve as Faraday cages by protecting occupants from electrical shocks..
In lighting systems, such as pendant lamps, the first equilibrium condition is used to fix them to the ceiling, floor or wall.
Objects placed on tables and shelves meet the first equilibrium condition. The normal force that the support exerts on the objects is responsible for compensating the weight.
To determine the viscosity of a liquid, a spherical object of known diameter is dropped into its interior and its speed will slow down due to resistance. The speed of the sphere is constant, thus being in dynamic equilibrium.
The higher the viscosity of the liquid, the lower the speed with which the sphere moves inside..
-Make a free-body diagram, showing all the forces that act on the body (omit those that the body exerts on others).
-Select a Cartesian coordinate system, ensuring that, as far as possible, the forces are located on any of the axes. The positive direction is usually taken in the direction of movement or a possible movement.
-Determine the Cartesian components of each force.
-Applying Newton's second law for each component, as established at the beginning, thus remains a system of equations.
-Solve the system of equations raised in the previous step.
The block of the figure, of mass m, it moves downhill on the inclined plane at angle θ with constant velocity. Calculate the value of the coefficient of kinetic friction μk, if the mass of the block is m = 5 kg and θ = 37º.
The first step is to draw the free-body diagram and choose a Cartesian coordinate system to vectorly express each force. The forces acting on the block are:
-The normal N exerted by the inclined plane, is perpendicular to the surface of this.
-The weight W is directed vertically down.
-Kinetic friction Fk that is opposed to movement. If it did not exist, the body would move downhill with an acceleration equal to g.senθ.
Like weight W is inclined with respect to the selected coordinate axes, it must be decomposed into its Cartesian components:
Wx = mg sin 37º = 5 kg x 9.8 m / stwo x sin 37º = 29. 5 N
WY = mg.cos 37º = 5 kg x 9.8 m / stwo x cos 37º = 39.1 N
Newton's second law is now applied, setting each summation equal to 0, since the block lacks acceleration when moving with constant velocity:
∑ FY = N - WY = 0
∑ Fx = Wx - Fk = 0
The magnitude of the kinetic friction is proportional to the magnitude of the normal, the coefficient of kinetic friction being μk the constant of proportionality.
Fk = μk N
At the same time:
N = WY = 39.1 N
Plus:
Fk = Wx
Therefore:
μk = 29. 5 / 39.1 = 0.75
Calculate the magnitude of the stresses that support the traffic light of mass 33 kg, shown in the figure:
The free-body diagram is made for both the traffic light and the knot that holds the cables:
Traffic light
On it act: the tension T3 up and W weight down. Therefore:
∑ FY = W - T3 = 0
Therefore:
T3 = 33 kg x 9.8 m / stwo = 323.4 N
Knot
The stresses are decomposed into their Cartesian components:
∑ FY = T1 sin 53º + Ttwo sen 37º - T3 = 0
∑ Fx = Ttwo cos 37º - T1 cos 53º = 0
And the following system of linear equations is obtained with two unknowns T1 and Ttwo :
- 0.6 T1 + 0.8 Ttwo = 0
0.8 T1 + 0.6 Ttwo = 323.4
The solution of this system of equations is: T1 = 258.7 N and Ttwo = 194.0 N
Equilibrium conditions.
Second equilibrium condition.
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